$f(x) = 4x^{2}-3x$ $h(t) = 2t^{2}+6t-5(f(t))$ $g(x) = -4x^{3}-6x^{2}-2x+6-f(x)$ $ f(g(1)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(1)$ . Then we'll know what to plug into the outer function. $g(1) = -4(1^{3})-6(1^{2})+(-2)(1)+6-f(1)$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = 4(1^{2})+(-3)(1)$ $f(1) = 1$ That means $g(1) = -4(1^{3})-6(1^{2})+(-2)(1)+6-1$ $g(1) = -7$ Now we know that $g(1) = -7$ . Let's solve for $f(g(1))$ , which is $f(-7)$ $f(-7) = 4(-7)^{2}+(-3)(-7)$ $f(-7) = 217$